\subsection{积分比大小}

	\begin{ti}
		设
		\begin{align*}
			a &= \iint_{D} \cos \sqrt{x^{2} + y^{2}} \dd{\sigma},\\
			b &= \iint_{D} \cos \bigl( x^{2} + y^{2} \bigr) \dd{\sigma},\\
			c &= \iint_{D} \cos \bigl( x^{2} + y^{2} \bigr)^{2} \dd{\sigma},
		\end{align*}
		其中 $D = \bigl\{ (x,y) \bigl| x^{2} + y^{2} \leq 1 \bigr\}$，则\kuo.

		\twoch{$c > b > a$}{$a > b > c$}{$b > a > c$}{$c > a > b$}
	\end{ti}

	\begin{ti}
		设平面区域 $D$ 由 $x = 0$，$y = 0$，$x + y = \frac{1}{4}$，$x + y = 1$ 围成，若 $I_{1} = \iint_{D} \bigl[ \ln (x+y) \bigr]^{3} \dd{x} \dd{y}$，$I = \iint_{D} (x + y)^{3} \dd{x} \dd{y}$，$I_{3} = \iint_{D} \bigl[ \sin(x + y) \bigr]^{3} \dd{x} \dd{y}$，则 $I_{1},I_{2},I_{3}$ 的大小顺序为\kuo.

		\twoch{$I_{1} < I_{2} < I_{3}$}{$I_{3} < I_{2} < I_{1}$}{$I_{1} < I_{3} < I_{2}$}{$I_{3} < I_{1} < I_{2}$}
	\end{ti}

	\begin{ti}
		设平面区域 $D = \bigl\{ (x,y) \bigl| (x - 2)^{2} + (y - 1)^{2} \leq 1 \bigr\}$，若比较 $I_{1} = \iint_{D} (x + y)^{2} \dd{\sigma}$ 与 $I_{2} = \iint_{D} (x + y)^{3} \dd{\sigma}$ 的大小，则有\kuo.

		\twoch{$I_{1} = I_{2}$}{$I_{1} > I_{2}$}{$I_{1} < I_{2}$}{不能比较}
	\end{ti}